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COMP20121 The Implementation and Power of Computer Languages Power - - PowerPoint PPT Presentation

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COMP20121 The Implementation and Power of Computer Languages Power Part http://www.cs.man.ac.uk/ petera/2121/index.html . Peter Aczel room: CS2.52, tel: 56155 petera@cs.man.ac.uk email: School of Computer Science, University of

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SLIDE 1

COMP20121 The Implementation and Power of Computer Languages ‘Power’ Part

http://www.cs.man.ac.uk/∼petera/2121/index.html .

Peter Aczel room: CS2.52, tel: 56155 email:

petera@cs.man.ac.uk

School of Computer Science, University of Manchester

COMP20121 - Section 0 – p.1/307

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SLIDE 2

COMP20121, ‘power’ part: section 1

lecture 1: Regular languages

LECTURE TWO

Finite automata for regular languages

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SLIDE 3

Match, or no match?

It can be hard to work out whether a given word matches a pattern.

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SLIDE 4

Match, or no match?

It can be hard to work out whether a given word matches a pattern. Consider the pattern: ab∗(a|b)b∗(a|b)∗,

COMP20121 - Section 1 – p.37/307

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SLIDE 5

Match, or no match?

It can be hard to work out whether a given word matches a pattern. Consider the pattern: ab∗(a|b)b∗(a|b)∗, and the word: abbababababbbababab. Does it match, or doesn’t it?

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SLIDE 6

Match, or no match?

It can be hard to work out whether a given word matches a pattern. Consider the pattern: ab∗(a|b)b∗(a|b)∗, and the word: abbababababbbababab. Does it match, or doesn’t it? We want to match letters in the target word with letters in the pattern. Answering ‘yes’ or ‘no’ is subject to a

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SLIDE 7

Automata

Instead of dealing with syntax, human beings often do better with pictures. e.g. when working with the pattern

(((0∗)|1)2|(0|(1∗))2) : you might draw

COMP20121 - Section 1 – p.38/307

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SLIDE 8

Automata

Instead of dealing with syntax, human beings often do better with pictures. e.g. when working with the pattern

(((0∗)|1)2|(0|(1∗))2) : you might draw

2 2 1 2 1

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SLIDE 9

Accepting a word

We can follow any word through the

  • automaton. Consider 0002.

2 2 1 2 1

COMP20121 - Section 1 – p.39/307

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SLIDE 10

Accepting a word

We can follow any word through the

  • automaton. Consider 0002.

2 2 1 2 1

COMP20121 - Section 1 – p.39/307

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SLIDE 11

Accepting a word

We can follow any word through the

  • automaton. Consider 0002.

2 2 1 2 1

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SLIDE 12

Accepting a word

We can follow any word through the

  • automaton. Consider 0002.

2 2 1 2 1

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SLIDE 13

Accepting a word

We can follow any word through the

  • automaton. Consider 0002.

2 2 1 2 1

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SLIDE 14

Accepting a word

We can follow any word through the

  • automaton. Consider 0002.

2 2 1 2 1

The double circle is an accepting state.

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SLIDE 15

Rejecting a word-I

We can follow any word through the

  • automaton. Consider 0001.

2 2 1 2 1

COMP20121 - Section 1 – p.40/307

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SLIDE 16

Rejecting a word-I

We can follow any word through the

  • automaton. Consider 0001.

2 2 1 2 1

COMP20121 - Section 1 – p.40/307

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SLIDE 17

Rejecting a word-I

We can follow any word through the

  • automaton. Consider 0001.

2 2 1 2 1

COMP20121 - Section 1 – p.40/307

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SLIDE 18

Rejecting a word-I

We can follow any word through the

  • automaton. Consider 0001.

2 2 1 2 1

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SLIDE 19

Rejecting a word-I

We can follow any word through the

  • automaton. Consider 0001.

2 2 1 2 1

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SLIDE 20

Rejecting a word-I

We can follow any word through the

  • automaton. Consider 0001.

2 2 1 2 1

Reject this word because we are stuck.

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SLIDE 21

Rejecting a word-II

We can follow any word through the

  • automaton. Consider 000.

2 2 1 2 1

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SLIDE 22

Rejecting a word-II

We can follow any word through the

  • automaton. Consider 000.

2 2 1 2 1

COMP20121 - Section 1 – p.41/307

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SLIDE 23

Rejecting a word-II

We can follow any word through the

  • automaton. Consider 000.

2 2 1 2 1

COMP20121 - Section 1 – p.41/307

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SLIDE 24

Rejecting a word-II

We can follow any word through the

  • automaton. Consider 000.

2 2 1 2 1

COMP20121 - Section 1 – p.41/307

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SLIDE 25

Rejecting a word-II

We can follow any word through the

  • automaton. Consider 000.

2 2 1 2 1

COMP20121 - Section 1 – p.41/307

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SLIDE 26

Rejecting a word-II

We can follow any word through the

  • automaton. Consider 000.

2 2 1 2 1

Reject: we end in a non-accepting state.

COMP20121 - Section 1 – p.41/307

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SLIDE 27

DFA—Definition

Definition 7 Let Σ be a finite alphabet. A (deterministic) finite automaton, (DFA),

  • ver Σ consists of (Q, q•, F, δ) where:
  • Q is a finite set of states;

COMP20121 - Section 1 – p.42/307

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SLIDE 28

DFA—Definition

Definition 7 Let Σ be a finite alphabet. A (deterministic) finite automaton, (DFA),

  • ver Σ consists of (Q, q•, F, δ) where:
  • Q is a finite set of states;
  • q• ∈ Q is the start state;

COMP20121 - Section 1 – p.42/307

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DFA—Definition

Definition 7 Let Σ be a finite alphabet. A (deterministic) finite automaton, (DFA),

  • ver Σ consists of (Q, q•, F, δ) where:
  • Q is a finite set of states;
  • q• ∈ Q is the start state;
  • F ⊆ Q is the set of accepting states;

and

COMP20121 - Section 1 – p.42/307

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DFA—Definition (continued)

  • δ is a transition function which for every

state q ∈ Q and every symbol x ∈ Σ returns the next state δ(q, x) ∈ Q.

  • So δ is a function from Q × Σ to Q.
  • When δ(q, x) = q′ we often write

q

x

→ q′.

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SLIDE 31

NFA—Definition

Definition 8 A non-deterministic finite automaton, (NFA), consists of (Q, q•, F, δ) where:

  • Q is a a finite set of states;

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SLIDE 32

NFA—Definition

Definition 8 A non-deterministic finite automaton, (NFA), consists of (Q, q•, F, δ) where:

  • Q is a a finite set of states;
  • q• ∈ Q is the start state;

COMP20121 - Section 1 – p.44/307

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SLIDE 33

NFA—Definition

Definition 8 A non-deterministic finite automaton, (NFA), consists of (Q, q•, F, δ) where:

  • Q is a a finite set of states;
  • q• ∈ Q is the start state;
  • F ⊆ Q is the set of accepting states;

and

COMP20121 - Section 1 – p.44/307

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SLIDE 34

NFA—Definition (continued)

  • δ is a transition relation which relates a

pair consisting of a state and a letter with a state.

COMP20121 - Section 1 – p.45/307

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SLIDE 35

NFA—Definition (continued)

  • δ is a transition relation which relates a

pair consisting of a state and a letter with a state. When (q, x) is δ-related to q′ we often write

q

x

→ q′ .

COMP20121 - Section 1 – p.45/307

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SLIDE 36

Variation of DFA: Dump states

Strictly speaking, our pictures do not quite fit the definition.

2 2 1 2 1

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SLIDE 37

Variation of DFA: Dump states

Strictly speaking, our pictures do not quite fit the definition.

2 2 1 2 1

There is no arrow labelled 1 from the red

  • state. So reject 001!

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SLIDE 38

Variation of DFA: Dump states

We do not always draw all the states. We can leave out ‘dump states’; i.e. states from which no accepting state can be reached. If we add a dump state to this DFA, it would look like this:

2 1 1 3 2 0, 1, 2, 3 0, 1, 2, 3 2 0, 3 1, 3

COMP20121 - Section 1 – p.46/307

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SLIDE 39

Variation of DFA: Dump states

We obtain this new picture by adding in one new state, . . .

2 1 1 2 2 0, 3

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Variation of DFA: Dump states

We obtain this new picture by adding in one new state, and arrows to this new state whenever there is an arrow missing in the

  • riginal picture.

2 1 1 3 2 0, 1, 2, 3 2 0, 3 1, 3

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SLIDE 41

Variation of DFA: Dump states

Further we add an arrow from the new state to itself labelled with all the letters of the underlying alphabet Σ = {0, 1, 2, 3}.

2 1 1 3 2 0, 1, 2, 3 0, 1, 2, 3 2 0, 3 1, 3

COMP20121 - Section 1 – p.46/307

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SLIDE 42

Variation of DFA: Dump states

If we draw all states, then we can follow a word through the automaton along a unique path without getting stuck. The word is accepted if and only if the state reached when the word ends is an accepting state.

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SLIDE 43

Variation of DFA: Dump states

If we leave out some dump states, then we can follow a word through the automaton along a unique path, but we might get

  • stuck. The word is accepted if and only if

we do not get stuck and the state reached when the word ends is an accepting state.

COMP20121 - Section 1 – p.46/307

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SLIDE 44

Accepting a word

Definition 9 A word α = x1 · · · xn is accepted by a DFA if

δ(q•, x1) = q1, δ(q1, x2) = q2,

. . .

δ(qn−1, xn) = qn ∈ F

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SLIDE 45

Accepting a word

This means that we have a sequence of states q•, q1, . . . , qn such that

q•

x1

→ q1

x2

→ q2

✲ · · ·

xn

→ qn ∈ F

COMP20121 - Section 1 – p.47/307

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SLIDE 46

Acceptance for NFAs

When we consider an NFA, there may be more than one way of following a word through the automaton. Consider the word

002.

2 1 2 1 2 0, 1

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SLIDE 47

Acceptance for NFAs

First attempt: Consider the word 002.

2 1 2 1 2 0, 1

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SLIDE 48

Acceptance for NFAs

First attempt: Consider the word 002.

2 1 2 1 2 0, 1

We get stuck and this is not an accepting path.

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SLIDE 49

Acceptance for NFAs

Second attempt: Consider the word 002.

2 1 2 1 2 0, 1

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SLIDE 50

Acceptance for NFAs

Second attempt: Consider the word 002.

2 1 2 1 2 0, 1

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SLIDE 51

Acceptance for NFAs

Second attempt: Consider the word 002.

2 1 2 1 2 0, 1

This is an accepting path and the word is accepted.

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SLIDE 52

Acceptance for NFAs

Third attempt: Consider the word 002.

2 1 2 1 2 0, 1

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SLIDE 53

Acceptance for NFAs

Third attempt: Consider the word 002.

2 1 2 1 2 0, 1

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SLIDE 54

Acceptance for NFAs

Third attempt: Consider the word 002.

2 1 2 1 2 0, 1

This is an accepting path and the word is accepted.

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SLIDE 55

Accepting a word—NFAs

Definition 9 ctd A word α = x1 · · · xn is accepted by an NFA if there are states

(q0 = q•), q1, . . . , qn

such that for all 0 ≤ i < n,

δ relates (qi, xi) with qi+1

and qn is an accepting state.

COMP20121 - Section 1 – p.49/307

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SLIDE 56

Accepting a word—NFAs

In other words, we have a sequence of states q•, q1, . . . , qn such that

q•

x1

→ q1

x2

→ q2

✲ · · ·

xn

→ qn

and qn is an accepting state.

COMP20121 - Section 1 – p.49/307

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SLIDE 57

Accepting a word—NFAs

For a DFA, it is easy to decide whether or not a given word is accepted. Just follow the unique path through the automaton given by the word.

COMP20121 - Section 1 – p.49/307

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SLIDE 58

Accepting a word—NFAs

For an NFA, there will typically be several paths given by the word. We have to investigate all of those in order to find out whether the word is accepted.

  • We will see later why NFAs are

nonetheless useful.

COMP20121 - Section 1 – p.49/307

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SLIDE 59

A DFA for a regular expression

Proposition 1.2 For every regular expression over some alphabet Σ there is a DFA that accepts precisely those words which match the regular expression.

COMP20121 - Section 1 – p.50/307

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SLIDE 60

A DFA for a regular expression

Proposition 1.2 For every regular expression over some alphabet Σ there is a DFA that accepts precisely those words which match the regular expression. However, this isn’t as easy as you might first think!

COMP20121 - Section 1 – p.50/307

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SLIDE 61

From patterns to DFAs,1

The obvious proof of Proposition 1.2 is to proceed by induction over the way a pattern is built.

COMP20121 - Section 1 – p.51/307

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SLIDE 62

From patterns to DFAs,1

The obvious proof of Proposition 1.2 is to proceed by induction over the way a pattern is built. And indeed, the start (base cases) looks promising.

COMP20121 - Section 1 – p.51/307

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SLIDE 63

From patterns to DFAs,2

We can define automata for the pattern ∅ (which matches no word at all).

Accepts no word at all

Σ

COMP20121 - Section 1 – p.52/307

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SLIDE 64

From patterns to DFAs,2

We can define automata for the patterns ∅ (which matches no word at all), ǫ (which matches precisely the word ǫ.

Accepts no word at all

Σ Σ

Accepts precisely ǫ

COMP20121 - Section 1 – p.52/307

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SLIDE 65

From patterns to DFAs,2

We can define automata for the patterns ∅ (which matches no word at all), ǫ (which matches precisely the word ǫ and x, which matches precisely the word x, where

x ∈ Σ.

Accepts no word at all

Σ Σ

Accepts precisely ǫ

x

COMP20121 - Section 1 – p.52/307

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SLIDE 66

The other forms of pattern

  • concatenation: (p1p2)
  • alternative: (p1|p2)
  • Kleene star: p∗

COMP20121 - Section 1 – p.53/307

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SLIDE 67

Concatenation

We start running into problems with concatenation:

A2 A1 A2 A2 A2 ? A1

+ =

COMP20121 - Section 1 – p.54/307

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SLIDE 68

Concatenation

Here is an example which shows why. The pattern a∗b∗ is the concatenation of a∗ with b∗.

b ¬b

+

a ¬a Σ Σ

COMP20121 - Section 1 – p.54/307

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SLIDE 69

Concatenation

If we treat this according to our original idea, we get

b ¬b

+

a b ¬b ¬a a ¬a Σ Σ Σ Σ

COMP20121 - Section 1 – p.54/307

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SLIDE 70

Concatenation

If we treat this according to our original idea, we get

b ¬b

+

a b ¬b ¬a a ¬a Σ Σ Σ Σ

But that allows the string aba to be accepted, which doesn’t match a∗b∗.

COMP20121 - Section 1 – p.54/307

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SLIDE 71

τ-transitions,1

The problem with the example clearly was that we couldn’t separate the two automata sufficiently.

b ¬b

+

a b ¬b ¬a a ¬a Σ Σ Σ Σ

COMP20121 - Section 1 – p.55/307

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SLIDE 72

τ-transitions,1

What we would have liked to do is something like this:

b ¬b a ¬a Σ Σ

COMP20121 - Section 1 – p.55/307

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SLIDE 73

τ-transitions,1

What we would have liked to do is something like this:

b ¬b a ¬a Σ Σ

We have drawn an arrow from every accept- ing state in the first automaton to the start state of the second automaton. Note that this arrow is not labelled!

COMP20121 - Section 1 – p.55/307

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SLIDE 74

τ-transitions,2

b ¬b a ¬a Σ Σ τ

The idea is that we should be able to take this transition whenever we like, without matching it against a letter of the word. We call such transitions τ-transitions and la- bel them in this way to avoid confusion.

COMP20121 - Section 1 – p.56/307

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SLIDE 75

Concatenation revisited

This allows us to deal with concatenation like this.

A1 A1 A2 A2

combine to

τ

COMP20121 - Section 1 – p.57/307

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SLIDE 76

Concatenation revisited

This allows us to deal with concatenation like this.

A1 A1 A2 A2

combine to

τ

Note that we need to append the automaton A2 (via a τ-move) for every accepting state of A1.

COMP20121 - Section 1 – p.57/307

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SLIDE 77

Concatenation revisited

So maybe we should be using this picture instead.

A2 A2 A2 A2 A1

+ =

A1 τ τ τ

COMP20121 - Section 1 – p.57/307

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SLIDE 78

Alternative

Using τ-moves makes it is easy to deal with alternative.

A1 A2

combine to

A1 A2 τ τ

COMP20121 - Section 1 – p.58/307

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SLIDE 79

Alternative

Using τ-moves makes it is easy to deal with alternative.

A1 A2

combine to

A1 A2 τ τ

Convince yourself that the new automaton recognizes the intended language!

COMP20121 - Section 1 – p.58/307

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SLIDE 80

Kleene star,1

The Kleene star leads to the most complicated looking automaton. This is because to be general enough to apply to all automata, we really do need all these

τ-transitions.

A

becomes

A τ τ τ τ

COMP20121 - Section 1 – p.59/307

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SLIDE 81

Kleene star,2

A

becomes

A τ τ τ τ

Convince yourself that the new automaton recognizes the intended language!

COMP20121 - Section 1 – p.60/307

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SLIDE 82

Kleene star,2

A

becomes

A τ τ τ τ

Convince yourself that the new automaton recognizes the intended language! For many examples we can make do with fewer states and τ-transitions than those given in the general rule!

COMP20121 - Section 1 – p.60/307

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SLIDE 83

Example 1

Consider the following automaton which corresponds to the pattern a|ab.

a b

COMP20121 - Section 1 – p.61/307

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SLIDE 84

Example 1

If we just try to insert τ-moves to get an automaton for (a|ab)∗, we get

a b τ τ τ τ

which accepts the word bb, which it shouldn’t.

COMP20121 - Section 1 – p.61/307

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SLIDE 85

Example 1

Let us now at least deal with the accepting states, as suggested in our rule.

a b τ τ τ τ τ τ

COMP20121 - Section 1 – p.62/307

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SLIDE 86

Example 1

a b τ τ τ τ τ τ

This does work, but if the start state was an accepting state as well, we might have to add another state.

COMP20121 - Section 1 – p.63/307

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SLIDE 87

Example 2

Let us assume we now want to find an NFA with τ-moves for the pattern a∗b∗.

COMP20121 - Section 1 – p.64/307

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SLIDE 88

Example 2

Let us assume we now want to find an NFA with τ-moves for the pattern a∗b∗. We start with the pattern a, which gives us the automaton

a

We apply the Kleene star to get an automaton for the pattern a∗.

τ τ τ τ a

COMP20121 - Section 1 – p.64/307

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SLIDE 89

Example 2

Let us assume we now want to find an NFA with τ-moves for the pattern a∗b∗. We apply the Kleene star to get an automaton for the pattern a∗. We can remove some superfluous τ-moves.

τ τ a

COMP20121 - Section 1 – p.64/307

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SLIDE 90

Example 2

Let us assume we now want to find an NFA with τ-moves for the pattern a∗b∗. We get an identical automaton for b∗, and putting the two together we obtain the solution to the overall problem.

τ τ τ τ τ b a

COMP20121 - Section 1 – p.64/307

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SLIDE 91

We have shown. . .

So what have we done now? We have constructed automata from patterns, but those automata mention a new kind of move (τ-transitions), and they are non-deterministic.

COMP20121 - Section 1 – p.65/307

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SLIDE 92

We have shown. . .

So what have we done now? We have constructed automata from patterns, but those automata mention a new kind of move (τ-transitions), and they are non-deterministic. In other words we have convinced

  • urselves that the following holds.

COMP20121 - Section 1 – p.65/307

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SLIDE 93

We have shown. . .

Proposition 1.3 For every regular expres- sion over some alphabet Σ there is an NFA with τ-transitions that accepts pre- cisely those words which match the regular expression.

COMP20121 - Section 1 – p.66/307

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SLIDE 94

From NFAs to DFAs,1

Our original aim was, of course, to construct a DFA (without any fancy moves like τ-transitions). We will therefore next concentrate on taking an NFA with such transitions and turning it into a DFA.

COMP20121 - Section 1 – p.67/307

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SLIDE 95

From NFAs to DFAs,2

In other words, we want to show the following result. Proposition 1.4 For every NFA with τ- moves there exists a DFA that accepts pre- cisely the same words.

COMP20121 - Section 1 – p.68/307